Jeeves prepares a hangover cure using four identical cocktail shakers.

Question

Jeeves prepares a hangover cure using four identical cocktail shakers. The first shaker is 1/2 full, the second shaker is 4/5 full, the third shaker is 1/k full and the last one is empty. After Jeeves redistributed all the content of the shakers equally into the four shakers, each shaker became 31/80 full. What is the value of k?

A. 2
B. 3
C. 4
D. 5
E. 6

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Bunuel · Accepted Answer

Official Solution:

Jeeves prepares a hangover cure using four identical cocktail shakers. The first shaker is 1/2 full, the second shaker is 4/5 full, the third shaker is 1/k full and the last one is empty. After Jeeves redistributed all the content of the shakers equally into the four shakers, each shaker became 31/80 full. What is the value of k?

A. 2
B. 3
C. 4
D. 5
E. 6

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Try easier version of the above question HERE.

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Attachment:

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A.  2 
B.  3 
C.  4 
D.  5 
E.  6

Let's denote the capacity of each shaker by  x .
 
Then, the amount of liquid in the first shaker is  \frac{1}{2}*x , the amount of liquid in the second shaker is  \frac{4}{5}*x , the amount of liquid in the third shaker is  \frac{1}{k}*x , and the amount of liquid in the fourth shaker is 0.
 
After redistributing the liquid equally among the four shakers, each shaker will contain  \frac{1}{4}^{th}  of the total amount of liquid, which is given to be  \frac{31}{80}*x , so we have:
  
 \frac{1}{4}(\frac{1}{2}*x + \frac{4}{5}*x + \frac{1}{k}*x) =\frac{31}{80}*x ;
 
 \frac{1}{4}(\frac{1}{2} + \frac{4}{5} + \frac{1}{k}) =\frac{31}{80}  (reduced by  x );
 
 \frac{1}{2} + \frac{4}{5} + \frac{1}{k} =\frac{31}{20}  (multiplied by 4);
 
 \frac{13}{10} + \frac{1}{k} =\frac{31}{20} ;
 
 \frac{1}{k} =\frac{31}{20}-\frac{13}{10}  ;
 
 \frac{1}{k} =\frac{1}{4} ;
 
 k =4 .

Answer: C

kj4325fd9s0 · Answer

k = 4
31/80 = 1/4 ( 1/2 + 4/5 +1/k)
Solve for k = 4

summerbummer · Answer

Didn't understand this.

I solved it using the following approach:

F/2 + 4F/5 + F/K = (31 x 4) / 80

Got F's value = 31 / 81

Then, F / K = 31 / 81
Substituted F's value and my ans was close to 2.

Bunuel  please help!

Bunuel · Answer

F, in your solution, denotes the capacity of one full shaker. In this case, it should be:

F/2 + 4F/5 + F/K = 31/80*4*F.

Solving the above will give k = 4 (F gets reduced).

IshanSaini · Answer

Lets consider that each shaker is of L ml.
so we have 4, L ml shakers.

If each of them was full, then we would have 4L ml of content in total.

Case 1 : now as per the question it is stated that :
shaker 1 = 1/2 L
shaker 2 = 4/5 L
shaker 3 = 1/k L
shaker 4 = 0 L

Case 2 : mind you, that after the redistribution of content each shaker is 31/80 full meaning :
shaker 1 = 31/80 L
shaker 2 = 31/80 L
shaker 3 = 31/80 L
shaker 4 = 31/80 L
or simple we could say that 4L was the total capacity and 31/80 of 4L will be quantity post redistribution.

Now the total content is same (in terms of ml) in case 1 and case 2, hence :
1/2 L + 4/5 L + 1/k L = 31/80 * 4L
->  solving this we get 1/k = 1/4, hence k = 4.

A. 2
B. 3
 C. 4 
D. 5
E. 6

NoraUnoSama · Answer

The contents of the 3 shakers divided by 4 will fill 31/80 of each of them and they're identical.

/4 = 31/80

= 31/80

You multiply both sides with 80

80/8 + 80/5 + 80/4K = 31

10 + 16 + 20/K = 31

20/K = 31 - 26 = 5

5K = 20

K = 20/5 = 4 = Answer C.

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Jeeves prepares a hangover cure using four identical cocktail shakers.

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